# Simplifying $\frac{1}{x} + \frac{5+x}{x+1} - \frac{7x^2 + 3}{(x+2)^2}$

I'm having trouble simplifying this expression:

$$\frac{1}{x} + \frac{5+x}{(x+1)} - \frac{7x^2 + 3}{(x+2)^2}$$

Would you first do the addition or subtraction?

What's the steps to solve this?

$$\frac{-6x^4 + 3x^3 + 26x^2 + 25x + 4}{x^4 + 5x^3 + 8x^2 + 4x}.$$

Thanks.

-
For some basic information about writing math at this site see e.g. here, here, here and here. –  Julian Kuelshammer Feb 7 '13 at 7:57

Or, if you set $I=\frac{1}x+\frac{x+5}{x+1}-\frac{7x^2+3}{(x+2)^2}$, then by multiplying $I$ by $x(x+1)(x+2)^2$, you get:

$$x(x+1)(x+2)^2\times I=x(x+1)(x+2)^2\left(\frac{1}x+\frac{x+5}{x+1}-\frac{7x^2+3}{(x+2)^2}\right)\\=(x+1)(x+2)^2+(x+5)x(x+2)^2-(7x^2+3)x(x+1)\\=3x^3+26x^2+25x+4-6x^4$$

-
Well done, of course!+1 –  amWhy Feb 8 '13 at 0:04

HINT

The guiding idea is the same as when you're evaluating $\frac{1}{3} + \frac{3}{4}$, which is to say that you find a common denominator. In my example, it would be $3\cdot 4$. In yours, it would be...

-
Would it be x * (x + 1)? –  Cypras Feb 7 '13 at 8:03
That's a good guess - but that's only the common divisor for the first two fractions. –  mixedmath Feb 7 '13 at 8:04
Yeah so I was wondering about that, is it better to do just two fractions at a time, or all three at once? –  Cypras Feb 7 '13 at 8:05
@Cypras: The idea above rules the fractions properly. +1 –  Babak S. Feb 7 '13 at 8:13
Common denominator would be x? –  Cypras Feb 7 '13 at 8:15